what does this whole sentence mean "It follows that we can now write I as a product of m transpositions in which the first transposition to be applied fixes n (this was proved under the assumption that τm(n) != n, and I is already in this form if τm(n) = n)." ?

what does this whole sentence mean "It follows that we can now write I as a product of m transpositions in which the first transposition to be applied fixes n (this was proved under the assumption that τm(n) != n, and I is already in this form if τm(n) = n)." ?

please help me understand lemma 1.4.3 with an example??

Thanks.....

I suggest getting more well-acquainted with $S_3$ which has 6 elements, and $S_4$ which has 24...the full symmetry groups get very big, fairly fast, and it becomes hard to write down everything explicitly for $n > 4$ (the multiplication table for $S_5$ has 14,400 entries).

What we want to do is prove the PARITY (evenness/oddness) of a transposition decomposition of the identity is invariant. The "number" of transpositions it takes to do so is NOT invariant for example, in $S_3$:

so we have a way to write the identity as a product of 2,4 or 6 transpositions right there.

Transpositions need not "start with a 1", in $S_4$ we have the following transpositions:

(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)

of which only the first 3 involve 1.

The first transposition only moves 1 and 2 (they swap places), so it fixes 3 and 4. Similarly (1 3) fixes 2 and 4, and so on.

*********

Ok, so we have an induction argument to carry out. Our base case is $n = 2$.

We assume that for some $n-1$ (maybe just 2, we don't know yet) the theorem is true.

We can thus regard our transpositions as elements of $S_{n-1}$. Now all of these can also be regarded as elements of $S_n$ that fix $n$.

Now we suppose we have some arbitrary product of transpositions in $S_n$ that equals the identity. Here's our strategy:

Suppose that the last transposition fixes $n$. Then we have a SHORTER product:

$\tau_1\cdots\tau_{m-1}$ that fixes $n$ as well. If $\tau_{m-1}$ ALSO fixes $n$, we keep going. If ALL of them fix $n$, then they all can be considered as lying in $S_{n-1}$, and we can apply our induction hypothesis to conclude that $m$ is even.

Of course, the assumption that $\tau_m(n) = n$ might not be TRUE. So here, we take a slightly different tack: we show we can re-write our product with DIFFERENT transpositions, in such a way that $m$ isn't changed.

So if $\tau_m(n) \neq n$ it must be that $\tau_m$ is of the form $(a\ n)$ for some number $a \in \{1,2,3,\dots,n-1\}$.

Next we look at the "next-to-last" transposition. There are 4 cases:

1. $\tau_{m-1}(n) = n$ and $\tau_{m-1}(a) = a$.

This implies that $\tau_{m-1}$ and $\tau_m$ are disjoint 2-cycles, so they commute, and we can just switch the order.

2. 1. $\tau_{m-1}(n) = n$ and $\tau_{m-1}(a) = b$ <--this is the same $a$ as we have in $\tau_m$.